\newproblem{lay:5_3_23}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.3.23}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	$A$ is a $5\times 5$ matrix with two eigenvalues. One eigenspace is three-dimensional, and the other eigenspace is two-dimensional. Is $A$ diagonalizable? Why?
}{
   % Solution
	According to Theorem 6.3.7, if the sum of the dimensions of the different eigenspaces is equal to the number of columns of $A$, then $A$ is diagonalizable. This
	is the case of the matrix $A$ of the problem for which $3+2=5$, and consequently $A$ is diagonalizable.
}
\useproblem{lay:5_3_23}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
